The reason that some of the elements of period three and beyond are steady in spite of not sticking to the octet rule is due to the fact of possessing the tendency of forming large size, and a tendency of making more than four bonds. For example, sulfur, it belongs to period 3 and is big enough to hold six fluorine atoms as can be seen in the molecule SF₆, while the second period of an element like nitrogen may not be big to comprise 6 fluorine atoms.
The existence of unoccupied d orbitals are accessible for bonding for period 3 elements and beyond, the size plays a prime function than the tendency to produce more bonds. Hence, the suggestion of the second friend is correct.
The answer is E and F.
Hope this helps!!
When calcium joins with calcium its follows the following equation:
2Ca + O2 = 2CaO
hope that helps
Answer:
(1) 2Al(s) + 2NaOH(aq) + 6H2O(l) ↔2Na[Al(OH)4](aq) + 3H2(g)
∴ Kc = ( PH2³ * [Na[Al(OH)4]² ) / [NaOH]² = 11
(2) H2O(l) + SO3(g) ↔ H2SO4(aq)
∴ Kc = [ H2SO4 ] / PSO3 = 0.0123
(3) 2P4(s) + 6O2(g) ↔ 2P4O6(s)
∴ Kc = Kc = 1 / PO2∧6
Explanation:
(1) 2Al(s) + 2NaOH(aq) + 6H2O(l) ↔ 2Na[Al(OH)4](aq) + 3H2(g)
∴ O / Al: 0 → +2 ≡ 2e-
Na: +1 → +2
∴ R / H: +1 → 0
2 - Al - 2
2 - Na - 1
8 - O - 8
14 - H - 14
⇒ Kc = ( PH2³ * [Na[Al(OH)4]² ) / [NaOH]² = 11
(2) H2O(l) + SO3(g) ↔ H2SO4(aq)
1 - S - 1
4 - O - 4
2 - H - 2
⇒ Kc = [ H2SO4 ] / PSO3 = 0.0123
(3) 2P4(s) + 6O2(g) ↔ 2P4O6(s)
8 - P - 8
12 - O - 12
⇒ Kc = 1 / PO2∧6
